# Solve for x, `log_10 (x^2 + 1) = 1`

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The equation `log_10(x^2+1) = 1` has to be solved for x.

For three positive numbers a, b and c, if `log_a b = c` then we can say `b = c^a` .

Given the equation `log_10 (x^2 +1) = 1` , as the base of logarithm is 10:

`x^2 + 1 = 10^1`

`x^2 + 1 = 10`

`x^2 = 9`

`x = +-3 `

The value x can take on both positive as well as negative values as x^2 is positive and so is x^2 + 1.

`log_10 (x^2 + 1) = 1`

The first step is to set this problem up in the exponential form:

`b^y=x`

we know that the y will b 1 because log is the exponent and the problem tells us log = 1

`10^1 = x^2 + 1`

simplify:

`10 = x^2 +1`

now subtract 1 from both sides to get `x^2` alone:

`9 = x^2`

now we take the square root of both numbers

`sqrt(9) = sqrt(x^2)`

`+- 3 = x`

to check:

`log_10 (3^2 + 1) =`

`log_10 (9 + 1) =`

`log_10 (10) = `

`log (x) / log(b) `

`log(10)/ log(10) = 1`

same goes for -3 because -3^2 is still 9